dc biasing bjts (analysis & design)es238/arquivos/referencias/...figure 1.2 Œ p2n2222 motorola...

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Visit http://electronicsclub.cjb.net for more resources DC BIASING BJTs (Analysis & Design)

• Design Procedure..(3) • Limits of operation.....(3) • BJT modes of operation...(4) • The Beta(hFE).....(5) • Experiment Determining the beta where it is stable.(6) • Analysis and design of dc-biased transistor configurations(9) • Fixed-Bias configuration...(9) • Effect of elements..(10) • Design Example 1 with electronics workbench.....(12) • Design Example 2 with ewb.(15) • Transistor Inverter , design example(3) with ewb ....(18) • Fixed-Bias circuit with RE....(20) • Effect of RE..(20) • Design Example 4 with electronics workbench.....(21) • Observation of stability between fixed-biased configuration and

fixed-biased with RE configuration..(28) • Voltage Divider Bias..(32) • Design Example 5 with electronics workbench.....(33) • Summary of voltage divider bias design..(36)


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Even if you are designing a transistor circuit as a switch or as an amplifier , transistor has to be biased in desired region. a.c purpose circuits are also designed according to DC conditions. So DC biasing is very important in both ac purpose(amplifier) circuits and DC purpose(switch) circuits.

There are some biasing configurations and in this tutorial these configuration will

introduced to you. Question: Why we dont use a typical bias configuration instead of many configurations? Answer: There are many biasing configurations , each one has advantageous and disadvantageous.Infact the main problem is stabilization.Some configurations are more stable when environment conditions are changed.


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DESIGN PROCEDURE

Biasing a BJT means establishing the desired values of VCE and IC so that the amplifier will have the proper gain, input impedance, undistorted output voltage swing, etc. These values of VCE and IC are known as the quiescent operating point or Q-point. The values of VCE and IC required are determined from inspection of the BJT's data sheet and load line analysis.

Whats the first step? Firstly limits of operations have to be known otherwise design procedure may be

more confusing.Generally limits of operations are obtained from the manufacturers datasheet.These datasheets can be found in manufacturers site.Some general purpose transistor datasheets can be obtained from http://electronicsclub.cjb.net. Limits of Operation The figure below illustrates a typical output characteristic of a transistor.The limits are often taken from the datasheet of the transistor Again , the maximum power dissipation is obtained from the datasheet and it is 300mW in this example.Note that , PCmax = VCE . IC .We can not use our transistor out of the PCmax limits.Operating point can be choosen at the points A,B & C.These points are at the active region of transistor.When amplification is intended , transistor have to work in active region and when switching operations are intended , the cut-off and saturation region is chosen for operating point of transistor.

Figure 1.1

After determining the limits ( PC max , VCE max , IC max ) , at least we will have an idea and know about the maximum values which must not be exceeded. Comments on point A,B and C: A : At lower IB levels spacing between IB curves is rapidly changing:Beta,hFE, is not

stable.Moreover negative swing is limited B : Best point for small signal amplification(Best symmetric swing occurs) C : Positive swing is limited


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BJT MODES OF OPERATION

*Forward biased ! VP > VN *Reverse biased ! VN > VP *For NPN transistors: Collector(n) , Base(p) , Emitter(n) * For PNP transistors: Collector(p) , Base(n) , Emitter(p) Active Region Condition: For NPN transistors: 1-Base-Collector junction must be reversed biased ( n>p ) , ( VC > VB ) 2-Base-Emitter junction must be forward biased ( p > n) , ( VB > VE ) VC > VB > VE VBE =0.7V For PNP transistors: 1-Base-Collector junction must be reversed biased ( n>p ) , ( VB > VC ) 2-Emitter-Base junction must be forward biased ( p > n) , ( VE > VB ) VE > VB > VC VEB =0.7V Cut-off region condition: Saturation region condition: VCE = 0.2V , VBE =0.8V VEC = 0.2V , VEB =0.8V For NPN Transistors :

Mode BE JUNCTION CB JUNCTION Cut-Off Reverse biased Reverse biased

Active Mode Forward biased Reverse biased Saturation Mode Forward biased Forward biased

For PNP transistors: VB > VE VB > VC

For NPN transistors: VE > VB VC > VB

For PNP transistors: VE > VB VC > VB

For NPN transistors: VB > VE VB > VC

VBE

VBC

Saturation

Forward-Active

Inverse-Active

Cuttoff


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The Beta (β or hFE)

Beta (hFE) is the D.C current gain. Its the ratio of the collector current to base current ! hFE = IC / IB ( IC = hFE. IB )

hFE is not a constant value it changes in different operating points.As IC increases or

decreases ( As well as IB ) hFE will change , but in somewhere it will be constant or more stable.Unless some special purpose circuits intented to design , we will bias our transistor in this beta stabilized region.

Figure 1.2 P2N2222 MOTOROLA DC CURRENT GAIN

Figure 1.3 BC546 DC CURRENT GAIN

As seen in the figure1.3 , the beta is fairly constant when IC is between 10-2 mA and 10mA.Now we have an idea about collector current in which interval it must be.Determining the collector current (IC) is one of the most important step.


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EXPERIMENT

DETERMINING THE REGION WHERE THE BETA ( hFE ) IS STABLE Our first experiment will be finding the region where the beta is more stable.In other words we will search where the beta is constant. Its better for you to practice these experiments in your home or lab.You can learn many things when doing an experiment.Moreover you may query yourself and try to make different configurations and see the results.This helps you to understand more clearly how things work. Even if you do not practice these exercises , read the theoritical informations and compare the measured results with the calculated ones. In this experiment the value of hFE ( βdc ) is examined and found where it is most stable.Two transistor is used to show you different results.(BC546 & 2N2222)

Figure 2.3 Q - Whats the saturation current? A - From equation (4) IC sat = VCC / RC

IC = 9V / 2.17KΩ ! IC sat = 4.14mA 2.2KΩ is measured as 2.17KΩ in this experiment.

Q - Whats the IB when RB is 470Ω A - From equaiton ( 2) IB = ( VCC 0.7V ) / RB IB = ( 9V 0.7V ) / 470Ω ! IB = 17.65mA Check the results with measured values in the next page Q - Whats the IC when IB is 5.55uA uA !microamper A - In order to answer this question we have to know the Beta (β) and if the the transistor works in active region. Lets give the answer of this question after the experiment results.

IB IC

VCC = 9V RC = 2.2KΩ ( Measured as 2.17KΩ its real value ) RB is changed from 470Ω to 20.5MΩ for obtaining different base currents IB and the response of the IC to the base current IB Q1:BC546 After using BC546 use 2N222 instead of BC546 Q1:2N2222


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Experiment Results (RC = 2.2KΩ and ICsat ≅ 4.14mA from equation (4) )

BC546 Q2N222 RB IB IC β ( hFE) # RB IB IC β ( hFE) 470Ω 17.66mA 4.13mA 0.233 1 470Ω 17.61mA 4.13mA 0.233 1K 8.3mA 4.16mA 0.5 2 1K 8.3mA 4.16mA 0.5 4.7K 1.76mA 4.16mA 2.36 3 4.7K 1.76mA 4.16mA 2.36 10K 0.836mA 4.16mA 4.97 4 10K 0.836mA 4.16mA 4.97 33K 0.258mA 4.16mA 16.1 5 33K 0.258mA 4.16mA 16.1 68K 0.122mA 4.15mA 33.7 6 68K 0.122mA 4.13mA 33.8 100K 83.7uA 4.15mA 49.5 7 100K 83.7uA 4.12mA 49.2 200K 41.95uA 4.14mA 98.7 8 200K 41.95uA 4.11mA 97.9 300K 28.03uA 4.14mA 147 9 300K 28.03uA 4.09mA 146 400K 21.02uA 4.13mA 196 10 400K 21.02uA 3.70mA 176 470K 17.9uA 4.13mA 230 11 470K 18uA 3.14mA 174 570K 14.5uA 4.11mA 283 12 570K 14.5uA 2.53mA 175 670K 12.35uA 4.09mA 331.7 13 670K 12.35uA 2.24mA 181 770K 10.81uA 4.08mA 378 14 770K 10.81uA 1.90mA 176 870K 9.64uA 4.04mA 420 15 870K 9.64uA 1.77mA 184 1M 8.11uA 3.51mA 433 16 1M 8.11uA 1.43mA 177 1.24M 6.78uA 2.97mA 438 17 1.24M 6.78uA 1.25mA 184 1.52M 5.55uA 2.46mA 443 18 1.52M 5.55uA 986uA 177 2M 4.21uA 1.83mA 435 19 2M 4.21uA 746uA 177 2.52M 3.35uA 1.5mA 449 20 2.52M 3.35uA 589uA 176 3.1M 2.72uA 1.22mA 448 21 3.1M 2.72uA 488uA 179 3.57M 2.37uA 1.07mA 453 22 3.57M 2.37uA 414uA 175 4.17M 2.03uA 912uA 449 23 4.17M 2.03uA 368uA 181 9.2M 0.923uA 391uA 424 24 9.2M 0.923uA 147uA 159 20.5M 0.413uA 193uA 468 25 20.5M 0.413uA 78.3uA 189

Table 1 When VC > VB > VE transistor is active region.In the measurements BC546 entered active mode at row # 15 , before # 15 it was in saturation region.The 2N2222 has entered active mode at row # 10. BC546 The β is fairly stable betwwen row # 15 and row # 24 , and so its β is determined as approximately 440 2N2222 The β is fairly stable betwwen row # 10 and row # 24 , and so its β is determined as approximately 180 Note that the saturation current is measured as approximately 4.16mA in both BC546 and 2N2222.It does not differ from transistor to transistor, saturation current is obtained from the equation (4) IC sat = VCC / RC IC sat = 9V / 2.17KΩ ! IC = 4.14mA The calculated value is nearly same as measured value that is ≅ 4.16mA As you see the answer of questions are very near to results that measured in the experiment.


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Now turning back to our last question ; Q - Whats the IC when IB is 5.55uA (row #18) uA !microamper A - From equation (3) IC = β . IB BC546 IC = β . IB , IC = 440 . 5.55uA ! IC =2.44mA ( Nearly same as measured value , look row # 18 ) 2N222 IC = β . IB , IC = 180 . 5.55uA ! IC =999uA ( Nearly same as measured value , look row # 18 ) Q - Whats the IC when IB is 1.76mA (row #3) A - From equation (3) IC = β . IB BC546 IC = β . IB , IC = 440 . 1.76mA ! IC =774mA ( ERROR! ) 2N222 IC = β . IB , IC = 180 . 1.76mA ! IC =316.8mA ( ERROR! ) The maximum current calculated and measured as ≅ 4.16mA , any value higher than ICsat =4.16mA can not exist! Equation (3) IC = β . IB is only applicable when transistor in active mode and where β is at its stable region Use this way , if equation (3) IC = β . IB satisfies or not : If IC (IC = β . IB ) is calculated higher than the saturation current IC sat ( IC sat = VCC / RC ) then the equation IC = β . IB is not applicable.Infact it is applicable but the β is lower than its stable value. Q Isnt it a problem , beta is not stable always? A Yes , it is a big problem.But there are some beta-independent configurations which will introduce to you later. So we give the name of the problem , Beta Problem!This problem will be removed in more improved configurations.


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ANALYSIS & DESIGN OF

DC-BIASED TRANSISTOR CONFIGURATIONS In this section some transistor configurations( DC-Biased) , thier designs & analysis , their advantages and disadvantages will be introduced to you.

1-FIXED-BIAS CIRCUIT

Figure 2.1

Figure 2.2 Fixed-Bias Configuration Kirchoffs Voltage Loops at input & output To get the neccessary equations we will use Kirchoffs voltage loops ( KVL ) at input & output From Loop-1 -VCC + IB .RB + VBE = 0 (1) VBE is 0.7 V IB = ( VCC 0.7V ) / RB (2) We have one more formula that is: IC = β . IB (3) (Active Region Equation) The equaiton (3) does not always satisfies IC = β . IB . It differs in saturation region. At Saturation: At saturation Vce ≅ 0.2V therefore KVL yields at output: IC = (VCC 0.2V) / RC (4)

Fixed-Bias is the simplest DC-biased transistor congfiguration.

Loop - 1 Input Loop

IB IC

Loop - 2 Output Loop


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Use this formula when transistor works in saturation region and keep the value of IB high enough to prevent your transistor work in active region , or in other words high value of IB keeps transistor in saturation region IBmax ≅≅≅≅ ICsat / β (The Level of IB in the active region just before the saturation) IB > ICsat / βmin (For the saturation level we must ensure this equation)

Three important equations for fixed-bias configuration:

EFFECT OF ELEMENTS EFFECT OF RB and IB : From equation (2) , IB is determined by RB : IB = ( VCC 0.7V ) / RB (2) As RB decreases IB increases , As IB increases Q point shift up-left as seen in the figure2.3.

Figure 2.3

IB = ( VCC 0.7V ) / RB (2) IC = β . IB (at active region) (3) At Saturation VCE=0.2V ; IC sat = ( VCC 0.2V )/ RC (4)

VCC

Q-point

Q-point

Q-point Q-point

Q-point RcVcc

IB1

IB2

IB3

IB4

IB5

RB IB Q-point shifts up-left RB IB Q-point shift down-right

Best Q point for small signal amp.Because it is at the middle of the graph.Therefore best symmetric swing occurs at this point.See ac analysis.

Higher the value of IB , more possible the transistor to be in saturation mode. Higher value of IB is obtained if lower value of RB is used

Adjusting RB for active mode and saturation mode operation , 1st Way Vcc Ic.RC = VCE (6).(at next page) IC = β.IB IBmax = 12V 0.8V(VCE) / β.RC (Maximum IB current when transistor works in acitve mode.)

Why 0.8V? VBE = 0.7V , VE = 0V ! VB = 0.7V VC > VB >VE .0.8V > 0.7V > 0V , VCE =VC = 0.8V border between active region and saturation region) IB = ( VCC 0.7V ) / RB (2) From Equation 2: RBmin = 12V 0.7V(VBE) / IBmax (Minimum RB , transistor to be work in active mode.)

If RB > RBmin or IB < IBmax then transistor will work in active mode.

If RB < RBmin or IB > IBmax then transistor will work in saturation mode


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EFFECT OF Rc: From loop-2 at page 5; IC = (VCC VCE ) / RC (6) This equation gives the DC load-line. Note that in active mode of operation IC is not determined by RC , IC is determined by equation (3) IC = β . IB. RC determines VCE=VC.See the example 1. For bigger Rc values , slope of the load line increases and Q-point shifts to left , this may limit the transistor operation in a smaller region.

Adjusting RC for active mode and saturation mode operation , 1st Way Max RC in active mode: In active region VC > VB > VE In fixed-biased configuration VE = 0V VBE = 0.7V therefore VB = 0.7V and VCE = VC IC = (VCC VCE ) / RC (6) RC = (VCC VCE) / IC (7) In active mode for max RC , VCE must be minimum and greater than 0.7V : VCE = VC > 0.7 V ( VC > VB ) ( VC > 0.7V ) Assume VCE = 0.8V RC = (VCC 0.8V) / β .IB (8) Maximum value of RC in active region

Q1 Q2 Q3

IB1

IB2

IB3

IB4

IB5 1RcVcc

2RcVcc

3RcVcc

RC1 > RC2 > Rc3 Changing Rc and keeping other variables constant shifts the Q point to the right or left on the same IB line.Ic remains constant if RC < RCmax Higher values of Rc may limit the transistor operation in a smaller region.

Adjusting RB for active mode and saturation mode operation , 2nd Way Active mode criteria : VC > VB > VE VC = VCC - ICRC VB = VCC IBRB VE = IERE

VC > VB ! VCC ICRC > VCC IBRB ! βIBRC < IBRB

!!!! RB > β.RC ( For RB > β.RC transistor works in active mode) RBmin = β.RC (Note that For RB < β.RC transistor work in saturation mode)

VB > VE ! VCC IBRB > IERE ! VCC > IERE + IBRB

! VCC >IB [β+1)RE + RB] (Transistor works in active mode) (Note that For VCC < IB [β+1)RE + RB] transistor work in saturation mode)


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Adjusting RC for active mode and saturation mode operation , 2nd Way The equation RB > β.RC can be used for adjusting the value of RC RC < RB / β ( For Rc < (RB / β) transistor works in active mode)

(Note that For Rc >( RB / β ) transistor work in saturation mode)

RC max = RB / β Minimum RC : VCC / RC = ICmax (9) RC = VCC / ICmax (10) Minimum vaue of RC ICmax can be found in the datasheet of the transistor.ICmax is illustrated in figure1.1 EFFECT OF Vcc

EXAMPLE 1: Vcc=12V , (βDC )hFE = 100

Solution:

a) Assuming transistor in active mode: VBE = 0.7V , VE = 0V ! VB =0.7V

IB = (12V 0.7V) / 240K ! IB = 47uA IC = β .IB IC = 100x47uA ! IC = 4.7mA IC = (VCC VCE ) / RC (6) From equation (6) : VCE = VCC IC.RC

IB1

IB2

IB3

IB4

IB5 VCC1 > VCC2 > VCC3

RcVcc1

RccVcc2

RccVcc3

Vcc3 Vcc2 Vcc1

Q2 Q1 Q3

Lower values of Vcc limit the transistor operation in a smaller region. Note that , as Vcc decreases IB also decreases.Therefore Q2 and Q3 points are in a lower IB level(not IB2 level , below IB2 level) This phenomenon is not illustared in the graph.

a) Determine IC , IB and VCE if RB=240K and RC=2.2K Determine the mode of operation

b) Determine the maximum and minumum value of RC

in active mode if RB=240K and ICmax=100mA c) Repeat a) for RB=240K and RC = 5K comment on the

result of a) d) Repeat a) for RB=240K and Rc =1K comment on the

result of a) and c)


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VCE = 12V (4.7mA).(2.2 K) ! VCE = 1.66V VE = 0V VCE = 1.66V ! VC = 1.66V VC = 1.66V VB = 0.7V VE = 0V VC > VB > VE Transistor is in active mode, our assumption is true.

Electronics Workbench Results for Example 1-a)

b) From equation (8) RC = (VCC 0.8V) / β .IB ( Maximum value of RC in active region )

RC = (12V 0.8V) / (100x47uA) ! RC max = 2.38K (Max value of RC in act. mode)

or RC < RB / β RC max = RB / β RC max = 240K / 100 RC max = 2.4K From equation (10) RC = VCC / ICmax ( Minimum value of RC)

RC = (12V ) / (100mA) ! RC = 120Ω (Min value of RC) IMPORTANT When RC smaller than 2.38K , transistor is in active mode and therefore IC is

constant and it is determined by equation (3) IC = β . IB. 120Ω < RC < 2.38K When RC greater than 2.38K , transistor will not be in active mode therefore IC is determined by equation (6) IC = (VCC VCE ) / RC RC > 2.38K

c) For RC = 5KΩ :We know transistor is not in active mode when RC > 2.38KΩ from b).

But for practicing we will again assume transistor in active mode(you will see now , it is not in active mode) Assuming transistor in active mode : VBE = 0.7 V VB = 0.7V IB = 47uA IC=4.7mA VC=VCE = VCC - IC.RC VCE = 12V (4.7mA).(5KΩ) !VCE=VC= -11.5 V VC < VB therefore our assumption is wrong!( transistor is not in active mode) Cut-off criteria: VE > VB VC > VB We know that VE = 0 and VB has a positive value VE < VB.Therefore transistor is not

in cut-off mode.We have only one choice that is : Transistor is in saturation mode. Saturation Criteria: VB > VE VB > VC VCE = 0.2V , VBE =0.8V

IB = (12V 0.8V) / 240K ! IB = 46.6uA IC = (VCC VCE ) / RC (6) IC = (12V 0.2V ) / 5KΩ ! IC = 2.36mA


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Electronics Workbench Results for Example 1-c)

IC is determined by RC only in saturation mode

d) For RC = 1KΩ :We know transistor is in active mode when 120 <RC < 2.38KΩ from b). But for practicing we will again assume transistor in active mode(you will see now , it is in active mode) Assuming transistor in active mode : VBE = 0.7 V VB = 0.7V IB = 47uA IC=4.7mA VC=VCE = VCC - IC.RC VCE = 12V (4.7mA).(1KΩ) !VCE=VC= 7.3 V VC > VB > VE therefore our assumption is true( transistor is in active mode)

As seen in part a) IC is same in both art a) and d) where RC < 2.38KΩ Remark:IC is not determined by RC in active region.It is determined by IC = β.IB

RC determines VCE , IC is determined by RC only in saturation mode.If RC high enough transistor works in saturation mode.

Electronics Workbench Results for Example 1-d)


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EXAMPLE 2: Vcc=12V , (βDC )hFE = 300

Solution:

a) Assuming transistor in active mode: VBE = 0.7V , VE = 0V ! VB =0.7V

IB = (12V 0.7V) / 240K ! IB = 47uA IC = β.IB IC = 300.47uA ! IC = 14.100mA VCE = VC = VCC IC.RC VC = 12V (14.100mA).(2.2K) ! VC = VCE = -19.02V VC < VB Therefore , transistor is not in active mode.Our asumption is wrong.We can determine whether transistor is in active mode or saturation mode by finding max RC as follows: max RC = (VCC 0.8V) / β .IB ( Equation 8 , max RC in active region.Note that 0.8 V comes from VC > VB , VB = 0.7 V , see notes) RC = (12V 0.8V) / (300x47uA) ! RCmax = 794ΩΩΩΩ (Max value of RC in active mode) or RC max = RB / β RC max = 240K / 300 RC max = 800Ω min RC : RCmin = VCC / ICmax (Equation 10 , min RC)

RC = (12V ) / (100mA) ! RC = 120Ω (Min value of RC) RC = 2.2K RC > RCmax 2.2K > 794Ω (Transistor in sat. Mode) if RC > RCmax then transistor works in saturation mode.

Assuming transistor in saturation mode:

Saturation Criteria: VB > VE VB > VC VCE = 0.2V , VBE =0.8V IC = (VCC VCE) / RC (Equation 4 , Saturation current , ICsat. IC≠β.IB ) IC = (12V 0.2V) / 2.2K ! Icsat = IC = 5.36mA

Consider the same circuit in example 1 , this time use β=300 in your solutions a) Determine IC , IB and VCE if RB=240K and

RC=2.2K.Determine the mode of operation and max RC and min RC.( take ICmax = 100mA)

b) Adjust RB so that transistor with Beta=300 and RC=2.2K will work in active mode.Determine IBmax and RBmin when RC is constant (2.2KΩ)


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Comment:Beta(β) affect the system if example 2 compared to example 1.When beta is 100 transistor works in active mode , when beta is 300 transistor works in saturation mode.To avoid saturation mode we have to use smaller RC (RC < 794Ω ), or smaller IB ( higher RB ).Note that smaller RC means higher VC and higher RB means lower VB.This settings adjust transistors mode of operation.See active mode and saturation mode criteria)


b) Vcc Ic.RC = VCE (6) From Equation 6:

IC = β.IB IBmax = 12V 0.8V(VCE) / β.RC (Maximum IB current transistor to be work in

acitve mode) IB = ( VCC 0.7V ) / RB (2) From Equation 2:

RBmin = 12V 0.7V(VBE) / IBmax (Minimum RB , transistor to be work in active

mode

IBmax = 12V 0.8V / 300.(2.2K) ! IBmax = 16.97uA RBmin = 12V 0.7V / IBmax RBmin = 12V 0.7V / 16.97uA ! RBmin = 665K or

RBmin = β.RC

RBmin = 300.(2.2K) RBmin = 660K if we use RB > 665K transistor will work in active mode if we use RB < 665K transistor will work in saturation mode Use factor of safety and do not choose the values near the limits


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For RB > RBmin , transistor will work in active mode RB = 700K ( 700 K > 665K ),

IB = (12V 0.7V) / 700K ! IB = 16.14uA ( Note that IBmax = 16.97uA ) IC = β.IB IC=(300).(16.14uA) ! IC = 4.842mA ( Note that ICsat = 5.36mA )

Electronics Workbench Results for Example 2-b) for RB > RBmin

For RB < RBmin , transistor will work in saturation mode RB = 600K ( 600 K < 665K ),

IB = (12V 0.7V) / 600K ! IB = 18.83uA ( Note that IBmax = 16.97uA ) IC ≠ β.IB

IC = (VCC VCE) / RC (4) IC = (12V 0.2V) / 2.2K ! Icsat = IC = 5.36mA

Electronics Workbench Results for Example 2-b) for RB < RBmin


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EXAMPLE 3: TRANSISTOR INVERTER

a) Determine RB and RC for the transistor inverter of figure above if ICsat = 10mA , β=300 b) Determine maximum value of RC for transistor to be work in active mode if RB=155K .

SOLUTION: a) At saturation IC sat = VCC VCE / RC (VCE =0.2V at saturation) IC sat = (Vcc 0.2V ) / RC 10mA = (10V 0.2V ) / RC ! RC = 980Ω At saturation : IB ≥ IC sat / βmin Assume β = βmin = 300 IB ≥ 10mA / 300 ! IB ≥ 33.3uA Choosing IB = 60uA to ensure that saturation. IB = (Vi 0.7V) / RB RB =(10V -0.7V) / 60uA RB = 155K


VC

t

10V 10V

0V

Vi

t

10V

0V 0V


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b) IB = (Vi 0.7V ) / 155K !IB = 60uA RC max = (VCC 0.8V) / β .IB (8) Maximum value of RC in active region RC max = (10V 0.8V)/18mA !RC max = 511Ω For RC < RC max , transistor work in active mode For RC > RC max , transistor work in saturation mode another way for max RC: RC max = RB / β RC max = 155K / 300 RC max = 516Ω


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FIXED-BIAS CIRCUIT with RE

Note that RE is reflected back to the input by a factor ( β +1 ) in active region. Effect of elements are same with fixed bias circuit. Effect of RE RE improves stability. RE determines VE RE affect base current much more than RB. As RE increases IB will decrease As RE increases slope of dc load line will increase(same effect as RC in fixed-bias configuration)Q point shifts down and left on the dc load line. RC and RE dont determines the IC in active mode.IC =β. IB At saturation mode IC sat = ( VCC 0.2V )/ RC + RE (2.3) Adjusting RB for active mode and saturation mode operation Active mode criteria : VC > VB > VE VC = VCC - ICRC VB = VCC IBRB VE = IERE

VC > VB ! VCC ICRC > VCC IBRB ! βIBRC < IBRB

!!!! RB > β.RC ( For RB > β.RC transistor works in active mode) RBmin = β.RC (Note that For RB < β.RC transistor work in saturation mode)

VB > VE ! VCC IBRB > IERE ! VCC > IERE + IBRB

! VCC >IB [β+1)RE + RB] (Transistor works in active mode) (Note that For VCC < IB [β+1)RE + RB] transistor work in saturation mode)

From input Loop: Vcc = IB.RB + VBE + IE.RE Vcc = IB.RB + VBE + (β+1)IB.RE At Active Region:

EB

BEB RR

VVccI)1( ++

−=

β (at active region) (2.1)

IC = β . IB (at active region) (2.2) From output Loop: Vcc = ICRC + VCE + IERE At Saturation Region: (VCE=0.2V ) IC sat = ( VCC 0.2V )/ RC + RE (2.3)

B

EBEB R

RIcsatVVccI

.−−= (2.4) Additional Eqns:

VC = VCC - ICRC VB = VCC IBRB VE = IERE


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Adjusting RC for active mode and saturation mode operation Active mode criteria : VC > VB > VE VBE = 0.7V IC = (VCC VCE ) / RC + RE (This equation gives the DC load-line.) VC > VB : VCE > VBE VBE =0.7V In active mode for max RC , VCE must be minimum and greater than 0.7V : VCE > 0.7 V Assume VCE = 0.8V

EB

C RI

VVccR −−=β

8.0max (Equation for determining max value of RC)

Note that aim of adjusting RC is changing the voltage VC The equation RB > β.RC can also be used for adjusting RC RC < ( RB / β ) ( For Rc < (RB / β) transistor works in active mode) RC max = RB / β (Another equation for determining max value of RC)

(Note that For Rc >( RB / β ) transistor work in saturation mode) The equation RB > β.RC can be used in either adjusting RB(keep RC constant) or adjusting RC ( keep RB constant) for determining the mode of operation. Minimum RC : VCC / RC = ICmax RC = VCC / ICmax (Minimum vaue of RC) ICmax can be found in the datasheet of the transistor.ICmax is illustrated in figure1.1 at page 3 EXAMPLE 4:

a) Assuming transistor works in active mode :

Active Region Criteria: VC > VB > VE VBE =0.7V IC = β.IB

EB

BEB RR

VVccI

)1( ++−

=β (2.1)

IB = [ 12V 0.7V ] / [ 470K + (301).(1.2K) ] ! IB = 13.59uA IC = β.IB IC = 300.(13.59uA) ! IC = 4.077mA

β = 300 , RB = 470K , RC =2.2K , RE =1.2K , VCC=12V

a) Determine mode of operation , IC and IB b) Adjust RB so that transistor works in active

mode.Whats the minumum value of RB if transistor wanted to be work in active mode?

c) Adjust RC so that transistor works in active mode.Whats the maximum value of RC if transistor wanted to be work in active mode?


22

VC = VCC IC.RC VC =12V (4.077mA).(2.2K) ! VC =3.03V VB = VCC IB.RB VB =12V (13.59uA).(470K) ! VB =5.61V VB > VC therefore transistor does not work in active region , our assumption is wrong

Assuming transistor works in saturation mode :

Saturation Mode Criteria: VB > VE VB > VC VCE = 0.2V , VBE =0.8V

IC sat = ( VCC 0.2V ) / RC + RE (2.3) ICsat = ( 12V 0.2V ) / (2.2K + 1.2K ) ICsat = 3.47mA

B

EBEB R

RIcsatVVccI

.−−= (2.4)

IB = [12V 0.7V (3.47mA)(1.2K)] / 470K !IB =15.18uA

Electronics Workbench Results for Example 4-a) b) Adjusting RB for active mode operation , 1st Way: ICsat = 3.47mA IBmax =ICsat / β IBmax = 3.5mA / 300 !IBmax = 11.56uA Assumption: IB have to be smaller than IBmax for transistor to be work in active mode.Check at the end if this assumption is true. IB < IBmax IB < 11.56uA

EB

BEB RR

VVccI

)1( ++−

=β (2.1)


23

Finding RB min:

11.56uA = [ 12V 0.7V ] / [ RB + (301).(1.2K) ] ! RB min = 616.3K

Any value of RB greater than RB min will cause transistor to work in active mode(You must check if this assumption is true) Taking the value of IB = 11.00uA

11uA = [ 12V 0.7V ] / [ RB + (301).(1.2K) ] ! RB = 666K

RB > RB min 666K > 616.3K Checking the assumption: IB = 11uA for RB = 666K IC = β.IB IC = 300.(11uA) ! IC =3.3mA

VC = VCC IC.RC VC =12V (3.3mA).(2.2K) ! VC =4.74V VB = VCC IB.RB VB =12V (11uA).(666K) ! VB =4.67V

VC > VB therefore our assumption is true.We can use RB=666K for active mode operation of transistor. Note that VC = 4.74V and VB = 4.67V are very close to each other.Any small difference in the system may cause the transistor to work in saturation mode.To avoid from this situation keep RB high enough as compared to RBmin.For example if we use RB = 800K , the difference between VC and VB will be more distinct then small changes will not change the mode of operation.

Electronics Workbench Results for Example 4-b) for RB = 666K Now we will investigate the truth of our assumption if we choose RB more close to RBmin . Taking the value of RB = 620K which is more close to RB min = 616.3K

IB = [ 12V 0.7V ] / [ 620K + (301).(1.2K) ] ! IB = 11.51uA IC = β . IB IC = (300)(11.51uA) ! IC = 3.45mA VC = VCC IC.RC VC =12V (3.45mA).(2.2K) ! VC =4.41V VB = VCC IB.RB VB =12V (11.51uA).(620K) ! VB =4.863V VC < VB therefore our assumption is wrong!

This way is not reliable when choosing values close to limits( For example RBmin = 616K RB=620K , the values are very close)

Use this way if you choose the value of RB distinct away from RBmin ( For example RB=800K)


24

Adjusting RB for active mode operation , 2nd Way: Active mode criteria : VC > VB > VE VC = VCC - ICRC VB = VCC IBRB VE = IERE

VC > VB ! VCC ICRC > VCC IBRB ! βIBRC < IBRB ! RB > β.RC VB > VE ! VCC IBRB > IERE ! VCC > IERE + IBRB ! VCC >IB [β+1)RE + RB] RB min = (300)(2.2K) RB min = 660K ( Actual result for minimum value of RB ) RB > RB min for active mode of operation Choosing RB = 670 K IB = [ 12V 0.7V ] / [ 670K + (301).(1.2K) ] ! IB = 10.95uA IC = β . IB IC = (300)(10.95uA) ! IC = 3.28mA VC = VCC IC.RC VC =12V (3.28mA).(2.2K) ! VC =4.784V VB = VCC IB.RB VB =12V (10.95uA).(670K) ! VB =4.66V VC > VB VE ≅ IC.RE VE ≅ (3.28mA)(1.2K) ! VE = 3.936 V VB > VE

VC > VB > VE : Transistor works in active mode It is an expected result because when RB = 666K transistor works in active mode as it is done in the previous page.if transistor works in active mode when RB = 666K , it will work in active mode for higher RB

Electronics Workbench Results for Example 4-b) for RB = 670K


25

c) Which IB will be used in the equation below?We can not use the IB which was was found IB =15.18uA in part a).Because IB=15.18uA is the base current when transistor is in

saturation mode.We need IB current when it is in active region.Because, if the equaiton below is investigated β.IB is written for IC.We know that β.IB is true for only in active region.

EB

BEB RR

VVccI

)1( ++−

=β (2.1) ( IB at active region)

IB = [ 12V 0.7V ] / [ 470K + (301).(1.2K) ] ! IB = 13.59uA

EB

C RI

VVccR −−=β

8.0max

(Any value of RC smaller than RCmax will not affect IC and transistor will work in active region ) RCmax = [(12V 0.8V) / 300.(13.59uA)] 1.2K RCmax = 1.547K For RC smaller than 1.547K make transistor to work in active region or 2nd way ( more easy) RC max = RB / β RC max = 470K / 300 RC max = 1.567K Choosing RC = 1K ( RC < RCmax ) and investigating the system: Assuming transistor works in active mode : Active Region Criteria: VC > VB > VE VBE =0.7V IC = β.IB

EB

BEB RR

VVccI

)1( ++−

=β (2.1)

IB = [ 12V 0.7V ] / [ 470K + (301).(1.2K) ] ! IB = 13.59uA IC = β.IB IC = 300.(13.59uA) ! IC = 4.077mA VC = VCC IC.RC VC =12V (4.077mA).(1K) ! VC =7.92V VB = VCC IB.RB VB =12V (13.59uA).(470K) ! VB =5.61V VE ≅ IC.RE VE =(4.077mA).(1.2K) ! VE =4.89V VC > VB > VE therefore transistor works in active region


26

Electronics Workbench Results for Example 4-c) for RC < RC max RC=1K , RC max = 1.547K

Choosing RC = 2K ( RC > RCmax ) and investigating the system: We know transistor works in saturation mode because RC > RC max.But we will try to solve the problem assuming it is in active mode.It will be seen that our assumption will be wrong. Assuming transistor works in active mode : Active Region Criteria: VC > VB > VE VBE =0.7V IC = β.IB

EB

BEB RR

VVccI

)1( ++−

=β (2.1)

IB = [ 12V 0.7V ] / [ 470K + (301).(2K) ] ! IB = 10.54uA IC = β.IB IC = 300.(10.54uA) ! IC = 3.162mA VC = VCC IC.RC VC =12V (3.162mA).(2K) ! VC =5.676V VB = VCC IB.RB VB =12V (10.54uA).(470K) ! VB =7.04V VC < VB therefore transistor does not work in active region.Our assumption is wrong as it was foreseen Assuming transistor works in saturation mode: Saturation Mode Criteria: VB > VE VB > VC VCE = 0.2V , VBE =0.8V IC sat = ( VCC 0.2V ) / RC + RE (2.3) ICsat = ( 12V 0.2V ) / (2K + 1.2K ) ICsat = 3.68mA

B

EBEB R

RIcsatVVccI

.−−= (2.4)

IB = [12V 0.7V (3.68mA)(1.2K)] / 470K !IB =14.64uA


27

VC = VCC IC.RC VC =12V (3.68mA).(2K) ! VC =4.64V VB = VCC IB.RB VB =12V (14.64uA).(470K) ! VB =5.12V

Electronics Workbench Results for Example 4-c) for RC > RC max RC=2K , RC max = 1.547K


28

OBSERVATION OF STABILITY BETWEEN FIXED-BIASED CONFIGURATION

AND FIXED-BIASED WITH RE CONFIGURATION

Outside conditions such as change in temparature , beta and age of the device will affect the system.The two fixed-biased configuration with RE and without RE are compared when β=50 and β=100(100% change in beta) Fixed-Biased Configurations without RE:

For β = 50

For β = 100

β IB(µµµµA) IC(mA) VCE(V) 50 20.43 1.026 6.865 100 20.54 2.049 1.64

β = 50 ! β = 100 ! 100% increase in IC


29

Fixed-Biased Configurations with RE = 2K , VE = 553.5mV :

For β = 50

For β = 100


β = 50 ! β = 100 ! 4% decrease in IB ! 95% increase in IC (improved stability)


30

Fixed-Biased Configurations with RE = 2K , VE = 1.790 :

For β = 50

For β = 100


β = 50 ! β = 100 ! 15% decrease in IB ! 76% increase in IC (The higher VE the more stable system) As seen in the results, emitter resistor RE improves the stability.Infact the stability is affected from VE.As VE increases the stability also increases.Note that VE can not exist if RE does not exist.Replacing a diode for RE for increasing VE will not make the system stable. We can generalize the subject and we can say that stability increases as VE increases and there must be a resistor RE for stability. We can adjust VE by changing the RE.(RB can also change the VE in active mode.)


31

Stability is determined directly by VE and RE and indirectly by RB.Observe the results. RE = 2K VE = 553.5mV ! 95% increase in IC when 100% increase in beta (less stable system) RE = 2K VE = 1.790mV ! 76% increase in IC when 100% increase in beta (more stable system)


32

VOLTAGE DIVIDER BIAS

Thevenin Equivalent :

The only purpose of RE is to provide DC feedback to stabilize the Q-point against variations in the β (also known as hFE) of the BJT. Since β can vary by as much as five to one from one BJT to another (even if the BJTs have the same 2N number), large unit-to-unit variations in Q-point could result unless this DC feedback is present. When the resistor values in the above figure are selected properly, β variations will have a negligibly small effect on the Q-point. To provide sufficient feedback, VE is usually chosen to be one-quarter to one-third of VCC. Because VBE =0.7 V (silicon devices), VE is set by the voltage divider formed by R1 and R2. However, reader will note from the above figure that R1 and R2 are not in series, and hence do not form a voltage divider. Only if the maximum value of IB is much less than I1(Current on R1) , so that IB can be neglected, will R1 and R2 act like a voltage divider. The circuit designer must establish this condition by the appropriate choice of values for R1 and R2.

This configuration is more stable then fixed-biased with RE configuration.IC and VCE will not affected by the change of beta if proper design is done. So we can call this configuration as Beta independent configuration. This configuration can be used for all BJT amplifiers (common emitter, common base, and common collector), although in the common-collector configuration we usually set RC = 0.


33

The effect of elements are again same as explained in past sections (Fixed-biased and fixed-biased with RE configuration).This time R1 and R2 will be explained in the solution of example 1. The design procedure is explained in the example 1.And a summary of design procedure will be found at the end of this tutorial. EXAMPLE 5:

SOLUTION: How we choose ICQ? IC is generally choosen by observing the datasheet of the transistor.Generally choosen at where β is maximum and stable. Note that in the datasheets IC - hfe graph depends on the voltage level of VCE.So choosen VCEQ have to be suitable for desired ICQ a) 1-Determinig the Q point:

For symmetric swing , 20.2V- Vcc VCEQ = (Check the datasheet if choosen ICQ is

suitable for calculated VCEQ ) VCEQ = (12V 0.2V) / 2 !VCEQ = 5.9V ICQ = 5mA , VCEQ = 5.9V gives the Q point on the load line. 2-Determine VE for stability: For stable systems: ( VCC / 10 ) ≤ VE ≤ ( VCC / 3 ) ( As VE increases , stability also increases ) Choosing VE = VCC / 4 VE = 12V / 4 ! VE = 3V 3-Determine RE: RE = VE / IC ( if IC≅ IE ) RE = 3V / 5mA ! RE = 600Ω

At IC=3 to 10mA maximum dc current gain occurs.Choosing IC = 5mA

a) Determine the value of the resistors for best symmetric swing at IC = 5mA

b) Instead of R2 ≤ (βmin.RE) / 10 (R2<< βmin.RE ) use R2 ≤ (βmin.RE) / 60 ( R2<<< βmin.RE)

VCC=12V β = 300


34

4-Determine RC: RC=VRC / IC VRC = VCC VCEQ VE VRC = 12V 5.9V 3V ! VRC = 3.1V RC = 3.1V / 5mA ! RC = 620Ω 5-Determine VB: VB = VE + 0.7V VB = 3V + 0.7V ! VB = 3.7V 6-Determine R2:

Ri = Equivalent resistance between base and ground looking into the base Ri = ( β +1 )RE ≅ βRE If Ri is much larger than the resistance R2 , the current IB will be much smaller than I2 and I2 will be approximately equal to I1.If we accept the approximation that IB is essentially zero amperes compared to I1 or I2 , then I1 = I2 and R1 and R2 can be considered series elements. Ri >> R2 βminRE ≥ 10R2

R2 ≤ (βmin.RE) / 10 Assuming β = βmin = 300 R2 = (300).600Ω / 10 ! R2 = 18K 7-Determine R1: The voltage across R2 , which is actually the base voltage , can be determined using the voltage-divider rule.

21

2B

VccR V

RR +=

3.7V = (18K).(12V) / (R1 + 18K) R1 =40.3K


35

Electronics Workbench results for example 1-a)

b) Same results are found until step 6 . RE = 600Ω , RC = 620Ω , VE =3V , VB = 3.7V , IC = 5mA , VCEQ = 5.9V 6-Determine R2: R2 ≤ (βmin.RE) / 60 (this time denominator is higher(10!60) that is R2 is more smaller) Assuming β = βmin = 300 R2 = (300).600Ω / 60 ! R2 = 3K 7-Determine R1: The voltage across R2 , which is actually the base voltage , can be determined using the voltage-divider rule.

21

2B

VccR V

RR +=

3.7V = (3K).(12V) / (R1 + 3K) ! R1 =6.72K

Electronics Workbench results for example 1-b)


36

As R2 and R1 gets smaller(I1 and I2 gets higher then IB has less effect on the system. ) the values for ICQ and VCEQ are more approximate to desired ICQ and VCEQ

SUMMARY OF VOLTAGE-DIVIDER BIAS DESIGN

1-Determinig the Q point:

Choose VCEQ as 20.2V- Vcc VCEQ = if symmetric swing desired.

Choose ICQ at the beta-stable region in the datasheet 2-Determine VE for stability: For stable systems: ( VCC / 10 ) ≤ VE ≤ ( VCC / 3 ) ( As VE increases , stability also increases ) 3-Determine RE: RE = VE / IC ( if IC≅ IE ) 4-Determine RC: RC=VRC / IC VRC = VCC VCEQ VE 5-Determine VB: VB = VE + 0.7V 6-Determine R2: R2 ≤ (βmin.RE) / 10 for more accurate design choose R2 smaller like: R2 ≤ (βmin.RE) / 30 7-Determine R1:

21

2B

VccR V

RR +=

dc biasing bjts (analysis & design)es238/arquivos/referencias/...figure 1.2 Œ p2n2222 motorola...

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